Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10166    Accepted Submission(s): 3179

Problem Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <queue>
 4 #include <cstring>
 5 using namespace std;
 6 int b,e;
 7 int Mintim;
 8 struct node
 9 {
10     int pos,t;
11 }k,tem;
12 int vis[100000+10];
13 queue<node> s;
14 void bfs()
15 {
16     while(!s.empty())
17         s.pop();
18     k.pos=b,k.t=0;
19     s.push(k);
20     while(!s.empty())
21     {
22         k=s.front();
23         s.pop();
24         if(k.pos==e)
25         {
26             Mintim=k.t;
27             return;
28         }
29         if(k.pos<0||k.pos>100000||vis[k.pos])    continue;
30         vis[k.pos]=1;
31         tem.t=k.t+1;
32         tem.pos=k.pos+1;
33         s.push(tem);
34         tem.pos=k.pos-1;
35         s.push(tem);
36         tem.pos=k.pos*2;
37         s.push(tem);
38     }
39 }
40 int main()
41 {
42     int i,j;
43     freopen("in.txt","r",stdin);
44     while(scanf("%d%d",&b,&e)!=EOF)
45     {
46         memset(vis,0,sizeof(vis));
47         bfs();
48         printf("%d\n",Mintim);
49     }
50     return 0;
51 }